Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $k = \dfrac{4r - 4}{2r^2 - 2} \times \dfrac{-4r^2 + 100}{2r + 10} $
Solution: First factor out any common factors. $k = \dfrac{4(r - 1)}{2(r^2 - 1)} \times \dfrac{-4(r^2 - 25)}{2(r + 5)} $ Then factor the quadratic expressions. $k = \dfrac {4(r - 1)} {2(r - 1)(r + 1)} \times \dfrac {-4(r + 5)(r - 5)} {2(r + 5)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {4(r - 1) \times -4(r + 5)(r - 5) } { 2(r - 1)(r + 1) \times 2(r + 5)} $ $k = \dfrac {-16(r + 5)(r - 5)(r - 1)} {4(r - 1)(r + 1)(r + 5)} $ Notice that $(r - 1)$ and $(r + 5)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {-16(r + 5)(r - 5)\cancel{(r - 1)}} {4\cancel{(r - 1)}(r + 1)(r + 5)} $ We are dividing by $r - 1$ , so $r - 1 \neq 0$ Therefore, $r \neq 1$ $k = \dfrac {-16\cancel{(r + 5)}(r - 5)\cancel{(r - 1)}} {4\cancel{(r - 1)}(r + 1)\cancel{(r + 5)}} $ We are dividing by $r + 5$ , so $r + 5 \neq 0$ Therefore, $r \neq -5$ $k = \dfrac {-16(r - 5)} {4(r + 1)} $ $ k = \dfrac{-4(r - 5)}{r + 1}; r \neq 1; r \neq -5 $